The energy scale of the qubit eigenvalues in D-Wave machine is GHZ *= 10^{-7}* Hartree
*= 10^{-5}* eV. This means that in our calculation 1 energy unit correponds
to *10^{-5}* eV or 100 mK.

Caculate the wave length of a phonon with energy *10^{-5}* eV in the following way:
The Debye energy is roughly *10^{-2}* eV. This corresponds to a wave vector *k=\pi
/a = 1*, where *a* is the lattice constant. The acoustic phonon has a linear
dispersion relation: *\omega = kv*, where v is the velocity of sound. Inserting
*\omega = 10^{-2}* and *k=1* yields *v=10^{-2}*. Therefore *k* at an phonon energy
of *10^{-5}* eV correponds to *k=10^{-3}*. The wave length is *\lambda = 2\pi
/k \approx 12\times 10^3bohr=10^4* Angstrom which is perhaps of the order of
the width of the Josephson junction. If instead we would require a much smaller
phonon wave length of e.g. 10 Angstrom, then *k=1* and the phonon energy would
be *\omega = 10^{-2}* eV. This is very large and phonons of this wave length cannot
be excited at a temperature of 100 mK.

D-wave operates below 100 mK and therefore phonons with a wave length shorter than
*10^4* Angstrom cannot be excited. Are the Mooij-Delft experiments an indication
that phonons of this wave length are nevertheless switching qubits?

When coupling to the environment the eigenvalues obtain an energetic width of *10^{-4}*
GHZ*= 10^{-9}* eV. If the environment would be phonons, they would have a wave length
of *10^8* Angstron comprising 10000 qubits. But phonons of such long wave length
can neither exist in the D-Wave machine, nor can they couple to a single qubit
and stabilize it against phonon excitations of shorter wavelength.

The time in the plots correspond to seconds, if energy unit is pico eV *= 10^{-12}*
eV. If energy units are interpreted as *10^{-5}* eV, then

10000 time units are 1 milli second

10 time units are 1 micro second

1 time unit is 100 nano seconds.